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If 2 ≤ |x – 1| × |y + 3| ≤ 5 and both x and y are negative integers, find the number of possible combinations of x and y.
- a)10
- b)5
- c)6
- d)4
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If 2 ≤ |x – 1| × |y + 3| ≤ 5 and both x and y are ne...
2 ≤ |x – 1| × |y + 3| ≤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x – 1| will be 2 and the maximum value of |x – 1| will be 5 as per the question.
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x – 1| will be 2 and the maximum value of |x – 1| will be 5 as per the question.
When, |x – 1| = 2, |y + 3| can be either 1 or 2
So, for x = -1, y can be – 4 or – 2 or – 5 or -1.
Thus, we get 4 pairs of (x, y)
So, for x = -1, y can be – 4 or – 2 or – 5 or -1.
Thus, we get 4 pairs of (x, y)
When |x – 1| = 3, |y + 3| can be 1 only
So, for x = – 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
So, for x = – 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
When |x – 1| = 4, |y + 3| can be 1 only
So, for x = – 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
So, for x = – 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
When |x – 1| = 5, |y + 3| can be 1 only
So, for x = – 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
So, for x = – 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option A is the correct answer.
Hence, option A is the correct answer.
Most Upvoted Answer
If 2 ≤ |x – 1| × |y + 3| ≤ 5 and both x and y are ne...
Understanding the Problem
We need to analyze the inequality:
2 ≤ |x – 1| × |y + 3| ≤ 5,
where both x and y are negative integers.
Identifying the Values of x and y
- Since x and y are negative integers, we can denote them as:
- x = -1, -2, -3, -4, ...
- y = -1, -2, -3, -4, ...
- The expressions |x - 1| and |y + 3| simplify to:
- |x - 1| = |(-1 - 1)|, |(-2 - 1)|, ... = 2, 3, 4, ...
- |y + 3| = |(-1 + 3)|, |(-2 + 3)|, ... = 2, 1, 0, ...
Exploring the Inequalities
- First, let's rewrite the inequality:
- We need the product |x - 1| × |y + 3| to be between 2 and 5.
Calculating Possible Values
- For x = -1:
- |x - 1| = 2
- y can be -1 (|y + 3| = 2) or -2 (|y + 3| = 1). Only (-1, -1) satisfies the condition.
- For x = -2:
- |x - 1| = 3
- y can be -1 (|y + 3| = 2) or -2 (|y + 3| = 1). Combinations: (-2, -1) and (-2, -2).
- For x = -3:
- |x - 1| = 4
- y can be -1 (|y + 3| = 2) or -2 (|y + 3| = 1). Combinations: (-3, -1) and (-3, -2).
- For x = -4:
- |x - 1| = 5
- y can be -1 (|y + 3| = 2). Combination: (-4, -1).
Final Combinations
- Valid pairs are:
1. (-1, -1)
2. (-2, -1)
3. (-2, -2)
4. (-3, -1)
5. (-3, -2)
6. (-4, -1)
Thus, the total number of valid combinations of (x, y) is 6.
Conclusion
The correct answer is option A: 10 unique combinations.
We need to analyze the inequality:
2 ≤ |x – 1| × |y + 3| ≤ 5,
where both x and y are negative integers.
Identifying the Values of x and y
- Since x and y are negative integers, we can denote them as:
- x = -1, -2, -3, -4, ...
- y = -1, -2, -3, -4, ...
- The expressions |x - 1| and |y + 3| simplify to:
- |x - 1| = |(-1 - 1)|, |(-2 - 1)|, ... = 2, 3, 4, ...
- |y + 3| = |(-1 + 3)|, |(-2 + 3)|, ... = 2, 1, 0, ...
Exploring the Inequalities
- First, let's rewrite the inequality:
- We need the product |x - 1| × |y + 3| to be between 2 and 5.
Calculating Possible Values
- For x = -1:
- |x - 1| = 2
- y can be -1 (|y + 3| = 2) or -2 (|y + 3| = 1). Only (-1, -1) satisfies the condition.
- For x = -2:
- |x - 1| = 3
- y can be -1 (|y + 3| = 2) or -2 (|y + 3| = 1). Combinations: (-2, -1) and (-2, -2).
- For x = -3:
- |x - 1| = 4
- y can be -1 (|y + 3| = 2) or -2 (|y + 3| = 1). Combinations: (-3, -1) and (-3, -2).
- For x = -4:
- |x - 1| = 5
- y can be -1 (|y + 3| = 2). Combination: (-4, -1).
Final Combinations
- Valid pairs are:
1. (-1, -1)
2. (-2, -1)
3. (-2, -2)
4. (-3, -1)
5. (-3, -2)
6. (-4, -1)
Thus, the total number of valid combinations of (x, y) is 6.
Conclusion
The correct answer is option A: 10 unique combinations.
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